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using analytical geometry,prove that the diagonals of a rhombus are perpendicular to each other please help me guys?????

 rhombus

Prove that Diagonals of a Rhombus are Perpendicular:

Consider a rhombus ABCD with AC and BD as diagonals.

To prove: AC and BD are perpendicular to each other

Let O be the point of intersection of the two diagonals.

Consider triangle ABO and BOC

AB = BC [AB and BC are sides of a rhombus. All sides of a rhombus are congruent.]

BO = BO [Common side]

CO = OA [The diagonals of a parallelogram bisect each other.]

By SSS (Side side side) theorm, we can say that triangle ABO is congruent to triangle BOC.

 

Since triangles AOB and BOC are congruent, all corresponding angles and sides should be equal. Hence we can say that angle AOB = angle BOC

Now we know that AC is a straight line, hence sum of angles AOB and BOC should be 180 degrees.

This can be expressed mathematically as:

Angle AOB + Angle BOC = 180

Since angle AOB = angle BOC, the above equation can also be rewritten as

Angle AOB + Angle AOB = 180 degrees

2 Angle AOB = 180

Angle AOB = 90 degrees

Angle BOC = Angle AOB = 90 degrees 

THUMBS UP PLEASE

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consider, a Rhombus ABCD with diagonals AC and BD intersecting at O. Clearly, AC and BD bisects each other

So, consider any two triangles, say AOD and COD so we have,

AD = CD, AO = CO and DO = DO (common) so, 

Triangles AOD and COD are congruent,

=> LAOD  = LCOD [c.p.c.t]

Now, LAOD + LCOD = 1800 [linear pair]

=> 2LAOD = 1800

=> LAOD = 900

=> LAOB = LCOB = LCOD = LAOD = 900 proved...

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In coordinate geometry ch. , how to prove....
does anyone know the full method...??
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using analytical geometry mot congurency 
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Let OABC be a rhombus such that OA is along x axis.Let BL and CM be prependiculars from B& C respectively on x-axis. Clearly, triangles ABL & OCM are congruent.

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